Solving Laplace $\nabla^2 \phi=0$ in $x,y \geqslant 0$ I'm trying to solve 
$\nabla^2 \phi=0$ in $x,y \geqslant 0$
$\phi(x,y)=0 $ as $x^2 +y^2 \rightarrow \infty$
$\phi_x(0,y)=0$ and $\phi(x,0)= \frac{1}{1+x^2}$
I know the solution is $\phi(x,y)=\frac{1+y}{x^2 +(1+y)^2}$
I want to use fourier cosine transform, so far I see the laplace equation becomes 
$\Phi_{yy}-k^2\Phi =0 $ where $\Phi$ is fourier transform in x. 
Then I get solution $\Phi(k,y)=A(k)e^{-|k|y}+B(k)e^{|k|y}$ 
I don't know what to do next
 A: The Fourier cosine transform on $x$ of Laplace's equation reveals that, 
$$\Phi(k,y)=A(k)e^{-ky}+B(k)e^{ky}$$
We know that $\phi(x,y) \to 0$ as $x^2+y^2 \to \infty$.  Since we are in the first quadrant only, this means that the solution vanishes as either $x$ or $y$ approaches $\infty$.  Thus, we have
$$\lim_{y \to \infty } \Phi(k,y)= \lim_{y \to \infty }\left(A(k)e^{-ky}+B(k)e^{ky}\right)=0$$
which implies that $B(k) =0$ and hence $\Phi(k,y)=A(k)e^{-ky}$.  We also know that $\phi(x,y=0) =\frac{1}{1+x^2}$.  Thus, we have 
$$\begin{align}
\phi(x,0) &=\frac{2}{\pi} \int_0^{\infty} \Phi(k,0) \cos (kx) dk \\
&=\frac{2}{\pi} \int_0^{\infty} A(k) \cos (kx) dk \\
&=\frac{1}{1+x^2}
\end{align}$$
which shows that
$$A(k) = \int_0^{\infty} \frac{1}{1+x^2} \cos (kx) dx = \frac{\pi}{2}e^{-k}$$
Finally, we have 
$$\begin{align}
\phi(x,y)&=\frac{2}{\pi} \int_0^{\infty} A(k)e^{-ky}\cos (kx) dk \\
&=\frac{2}{\pi} \int_0^{\infty} \left (\frac{\pi}{2}e^{-k}\right) e^{-ky}\cos (kx) dk \\
&=\text{Re} \left( \int_0^{\infty}  e^{-k(1+y-ix)} dk             \right) \\
&=\text{Re} \left( \frac{1}{(1+y)-ix}           \right) \\
&=\frac{1+y}{(1+y)^2+x^2}
\end{align}$$
